Casing Set Depth: 2017 #55

A well’s surface section is drilled to 8,000 ft TVD where the pore pressure is 3,000 psig and the fracture pressure is 6,000 psig. At 4,000 TVD the pore/fracture pressures are 1,500/4,000 psig, and 5,000/8,000 psig at 10,000 ft TVD.

Assuming a trip margin of 200 psi and a kick margin of 500 psi, and a future intention to drill to 10,000 ft TVD once the surface casing is set, the maximum mud weight allowed to drill the surface section is closest to:   (A) 14.1 lb/gal (B) 13.2 lb/gal (C) 13.5 lb/gal (D) 13.8 lb/gal

This is pretty simple once you've read through all the details. See 6 DTC 6 (csg set depth) and it has the single needed equation: Max MW=(Pff - KM)/0.052(TVDshoe) = (5,000-500)/(0.052*8,000) = 13.2 or (B). The GB has the exact problem as an example.

Casing Design: 2017 #54

A well’s production casing will be set at a TD of 16,000 ft in 10.2 lb/gal mud. Which statement is most FALSE? 

(A) Production casing design should start at 16,000 ft and move uphole. 
(B) Casing collapse resistance must be reduced for tension if below 13,500 ft. 
(C) Casing collapse resistance needn’t be reduced for tension if below 14,500 ft. 
(D) Casing collapse resistance must be reduced for tension for casing above 12,500 ft. 

This problem has a lot of garbage you don't need (I don't include it here). (A) is clearly true (csg design starts on bottom and moves uphole; look up as needed). (B) through (D), however, are really the same question regarding collapse in relation to depth.

Solution: See Guidebook 6 DTC 9 that covers casing design and the neutral plane. Next, look up the 10.2 ppg buoyancy factor to calculate the hole section's neutral plane (16M*0.844 = 13,500 ft). The rest is academic; as TS 12 puts it on P430: Collapse performance properties will require derating for tension above the neutral plane. So (B) is false. 

But watch the wording on these types of problems like a hawk. You might work 5 minutes then mess up the answer due to some double-negative in the wording, confusion over "above" versus "below" the NP, or if the text is actually "true" or "false", even if you understand the problem fully. For this reason I always save 30 seconds for a check/re-read of each problem. Paranoia here is your friend.

Separator: 1017 #53

Problem 53. What is most FALSE about oilfield separators: 

(A) Horizontal handles foam but not solids more efficiently than vertical... 
(B) SPEC 12J is the standard API reference for oil & gas separator design. 
(C) Gravity separation section: reduces entrained liquid load & improves gas velocity profile. 
(D) Souders-Brown approach for sizing the gravity separation section...is no longer recommended.

This is a general separator question; see 8 FAC 5. This page quickly shows that A & B are true. Say 90 seconds down.

General knowledge tells me C is true as well, leaving D as the answer; I mark that at 2 minutes.

However,  I prefer verification. I'm checking references; the Guidebook recommends HS3 P15-47 and 12J, so I go to HS3 first. After another two minutes I verify from the text that C is indeed true.

Now I'm on to D. I'm pretty sure it's the answer, but again I want to verify this and so check the HS Index for Soulder. Nothing. I check the Dictionary? Nothing. Bradley? Nothing! Next 12J, which I happen by luck to have (1999 edition)? Nothing! Out of desperation, I check Mian. Nothing!! I'm now past my 6 minutes, But I'm pretty sure it's D from process of elimination and I move on. I've wasted a lot of time I could have invested elsewhere.

But for those curious, basically 12J uses a Ks factor via the Souders-Brown equation as an empirical parameter (but never really tells us the how or why). As we have seen, it's is not obvious in any of the common references. Be prepared for this; the solution may not be in any book you have and you must use your experience, common sense, test-taking skills, and the process of elimination. Watch the clock like crazy when using references. As I often say: a person's chances of passing an open-book exam is inversely proportional to the number of pages referenced. And most of the time, it's just a waste of time.

Fishing: 2017 #48

Problem 48. This is a standard problem just like in the Guidebook on 6 DTC 10. Math below. Note how much quicker using the "sc" chart for the tubing and casing is than doing the entire FL/EA calculation each time:

10,000 lbf(8103’ dp)1.5444E-07 (sc for dp) = 12.5”.
20.5” – 12.5” = 8”.
8”/[(10,000 lbf)1.1047E-07] (sc for tbg) = 7,243’ (A).

2017 #46

Which specific operation would typically incur the largest ton-miles using a 1-1/8” drilling line (assume a typical US onshore rig, typical BHA and drillpipe, and the same well)? 

(A) Drilling from 4,000 to 5,000 ft using a 8.5 inch PDC bit. 
(B) A round trip at 5,000 ft. 
(C) Core from 5,000 ft to 5,080 ft. 
(D) Running 7 inch 26 lb/ft casing to 5,080 ft. 

See 1 RIG 5. After calculating RTTM multiply the following factors based upon the actions: 1) Trip/csg: x1; 2) Coring x2; 3) Drilling x3. This makes sense; drilling has 3x the drill line action due to tripping/wiper/drilling, coring 2x due to no wiper, and tripping is one-way.

But in this problem the multipliers are moot. A) 1,000 ft drilling, 2) 5,000 ft round trip or 10,000 ft, C) 80 ft core, running casing 5,080 ft. The round trip dominates all multipliers, or (B).

Reservoir Volume: 2017 #44

A single well is drilled in a small isolated volumetric oil reservoir with a total compressibility factor 0.00002 1/psi and an oil FVF of 1.3 bbl/STB.

The well flows at 100 BOPD until pseudosteady-state flow is reached and maintained until BHP falls 125 psi over 4 days. The reservoir volume is closest to (Mrb): (A) 420; (B) 320; (C) 220; (D) 120.

This is plug-and-chug using 12 WLT 3: -0.234qB/(ct dpwf/dt) = 0.234(100)1.3 / (2E-5(125/4*24), which at 5.615 cf/bbl converts to 208 Mrb, or (C).

Just be careful with the units, and know the provided CBT reference locations for unit lookup.

Rod Pump VE: 2017 #42

Problem 42. The volumetric efficiency of the lift system in Problem 41 above (19 bbl oil+ 188 bbl water) is closest to...

On 7 PRD 9, we see: PD = 0.1166(64)10(1.75^2) = 228 bpd (prior problem).
Volumetric Efficiency = (Produced Volume)/(Pump Displaced Volume) = 19 + 188 = 207 so:
VE = 207/228 = 91%.

Beam Counterbalance: 2017 #41

Problem #41: 25 bbl of 40 API oil and 188 bbl of water (SG = 1.1) are produced from a pumping unit operating with...The counterweight required is nearest to: (A) 9,900 lbs; (B) 10,100 lbs; (C) 10,300 lbs; (D) 10,500 lbs.

I've shown several examples of this type of problem on the blog. Why? They are quick, common rod pump problems because they don't require lengthy charts or tables. So expect them and know how to do them fast. Everything you need to solve this problem is in the Guidebook on two pages 7 PRD 8-9. Solution (note the corrections to the fluids above do not effect this problem's solution):

7/5 rods; 1.75” pump@5000: Wr = 1.732 lb/ft & L = 5000 (7 PRD 10).
W = Wr*L = 1.732(5,000 ) = 8,660 lbs.
Find G: (WC)SGw+(OC)SGo = (0.9)1.1+(0.1)0.825 = 1.07 (7 PRD 1).
Wrf = W(1-0.128G) = 8660(1-(0.128*1.07)) = 7,474 (7 PRD 9).
Fo = 0.34*1.07*1.75^2*4000 = 4,457.
CBE = 1.06(Wrf+(0.5*Fo)) = 1.06(7474+(0.5*4457)) = 10,285 lbs (C).

Gas Meters: 2017 #40

A gas meter orifice:
(A) Typically has less than 10% of an effect...
(B) Should be installed with the beveled edge upstream. 
(C) Should be installed with the sharp edge downstream. 
(D) Typically has less than 20% of an effect...

The Guidebook (8 FAC 3 & 4) states gas orifice plates are installed with beveled edge downstream. So we know B & C are false.

The same page also says an orifice installed backwards reads about 15% low. So we know A is false.

Finally, since we know backwards is the greatest error it can have due to orientation (15%), D must be true.

LOT: 2017 #37

A LOT at 10,000 ft with 10 ppg mud records the following volumes of fluid and pressures after pumping: 3 bbl, 600 psi; 4 bbl, 800 psi; 4.5 bbl, 900 psi; 5 bbl, 950 psi. If 100 psi was lost overcoming the mud gel strength, the formation fracture pressure is closest to:

This is the Guidebook example. It's also a fair question because it tests one's understanding of a LOT. One should have the needed page in a few seconds using the TOC under "LOT".

The hydrostatic pressure calculates to 5,200 psi, 900 psi is added, and 100 psi is subtracted making 6,000 psi. Simple and fast if you understand the process and have the equation. Just be prepared to graph it out on scratch paper.

Csg MW Max/Min: 2017 #36

For casing set depth:
MW max is limited by prior casing shoe fracture pressure (minus kick margin).
MW min is limited by BHP (plus trip margin).

In this problem:

3,000 psi = 0.052(MWmax)4,000 ft
...so MW max = 14.4 ppg

 6,000 psi = 0.052(MWmax)9,000 ft
 ...so MW max = 12.8 + 0.2 = 13.0 ppg

Well Communication: 2017 #35

Problem 35. Your company investigates a lease sale containing two wells, A and B. Kelly bushing elevations are 1,000 ft and 0 ft above MSL, respectively, and bottom hole pressures of 3,170 and 3,500 psi, respectively. Both wells are 11,000 ft vertical depth to the perforations with pressure gradients of 0.3 psi per foot. The seller claims the two wells are in communication. Which of the following is most likely TRUE?
 
(A) The seller is incorrect; the wells are about 6 percent different in reservoir pressure and unlikely to be in communication.
(B) The seller is correct; the two wells are about 1 percent different in reservoir pressure and thus likely to be in communication.
(C) The seller is incorrect; the wells are over 10 percent different in reservoir pressure and unlikely to be in communication.
(D) The seller is correct; the two wells are about 4 percent different in reservoir pressure and thus likely to be in communication.

This problem is pretty straightforward and calculates to "B". But it requires practical but diverse petroleum engineering knowledge that trips many up.

Specifically: how much pressure difference is there within a reservoir from well to well? How to calculate pressures in wells that have different MSL depths? Nothing complex, just meat-and-potatoes for the average experienced engineer, but it can take time to think it through and look for possible tricks.

If you have more questions (or notice an error) note it in the comments below. For more help, this type of problem is in 13 RES 8 and SPE TS8 (where I got the idea for it). Myself, I'm used to drilling in the flats...for Rocky Mountain types this sort of problem is probably a yawn...

Reciprocating Compressor VE: 2017 #32

This problem uses the standard reciprocating compressor volumetric efficiency equation (found in the Guidebook or the SPE Handbook Series). It can be difficult to recognize this, though, when you are merely asked for a gas rate.

It's fairly plug-n-chug once you have the necessary equation. The only trick? "L" (gas slippage) is not given so you must draw on experience assuming it's between 0-5%. When I did this problem I used 0%, then 5%, and found both gave the same answer "B": (A) 7.4 SCF/min (B) 8.4 SCF/min (C) 9.4 SCF/min (D) 6.4 SCF/min.

Displacement: 2017 #20

5,000' of 14.00 lb/ft drillpipe (adjusted weight is 1.1 times nominal) and 100 ft of 120 lb/ft drill collars are pulled dry. Fluid volume change?

14 ppf(1.1)5,000' =  77M lb.
120 ppf(100') = 12M lb.
89 M lb/(2751 lb/bbl) = 32 bbl (A).

The trick on this one is to use adjusted weight whenever given; never use OD & ID unless you don't have adjusted weight. Then use the adjusted weight / density equation (see 1 RIG 1). Know how to do this sort of problem fast.

Note some prefer to use 490 lbm/cf x 5.615 cf/bbl rather than 2,751 lb/bbl that I use here since they have 490 &amp & 5.615 memorized.

ESP: 2017 #19

ESPs have many testing options. In this problem, we are given:

1) SG of the working fluid (10% water SG 1.01, 90% oil SPI 40)
...so SG = 0.1(1.01) + 0.9(0.8244) = 0.84

2) Net Lift 3,500 ft, Friction 300 ft, Surface pressure 200 ft
...so TDH = 3,500 + 300 + 200 = 4,000 ft

3) ESP with head capacity 4,000 ft/100 stages and 1 BHP/stage:
...4,000 ft / (4000 ft/100 stages) = 100 stages
...so BHP = (1 BHP/stage)100 stages(0.84) = 84 BHP

Fairly simple. If you aren't familiar with all three steps before seeing the problem it's hard to do it quickly. So be familiar with all three steps.

Displacement: 2017 #18

This is a fairly easy problem. The annulus is 10,000 ft of 10 ppg mud which weighs in at 5,200 psi.

The drillpipe has three layers: 4,000 ft of 8 ppg (1,664 psi), 4,000 ft 9 ppg (1,872 psi), and 2,000 ft 10 ppg mud for a total 4,576 psi. Thus we need another 625 psi for pressure balance.

This problem is easy, but hey not every problem can be hard.

Oil Saturation: 2017 #17

Problem 17. The oil from a volumetric undersaturated reservoir has a formation volume factor is 1.4 bbl/STB. The water saturation is 19 percent. 6 percent of the oil is produced and the new FVF is 1.45. The remaining gas saturation percentage is closest to: (A) 0.1 (B) 1 (C) 5 (D) 25.

This is a simple plug-and-chug problem with the needed equation. However, it's not common. 13 RES 4 has the equations you need:
Oil saturation = (1 – 0.06)(1 – 0.19)(1.45/1.4) = 0.788
Gas saturation = 1 – So – Swi = 1 – 0.788 – 0.19 = 0.021, or 2.1%, or (B).

First, one needs to know the proper equation. Noting the givens: undersaturated reservoir, the percentage produced, and appropriate formation volume factors should clue all you in.

Next, the answer is calculated as decimal...yet it asks for percentage with the decimal as an option. Not nice.

Finally, it doesn't give the exact match but makes you wonder. Rare, but it can happen, so double checking your work at all times is all one can do.

Pipe Collapse: 2017 #14

The problem reads: Given 5-1/2” 26 lb/ft N80 casing with 40,000 psi tension: find the reduced collapse rating (if your older copy is missing pipe weight, please assume it):

The whole problem can be solved never leaving 6 DTC 4.
The 80 clues us to an max axial stress of 80,000 psi. Add the internal pressure to compute: (40,000 + 20,000)/80,000 = 0.75.
Enter this into the GB table (ellipse of plasticity) for -0.385.
Solve using GB equation (with collapse rating from HES Redbook of 12,650 psi) for: 12,650(0.385)+20,000 = 24,875 psi.

Diagenetic Porosity: 2017 #12

Porosity calculations are shown on a single GB page: 15 LOG 4. It starts:

Total Porosity: (measured with nuclear tools).
…equals primary porosity + secondary porosity.

Primary Porosity (apparent, intergranular).
Found from Wylie’s (acoustic) porosity equation below.

Secondary Porosity (isolated pores, vugs, and fractures).
also called diagenetic porosity.
…may be overlooked by acoustic-logs.
…equals total porosity – primary porosity.

At a glance, it's easy to see diagenetic porosity is found by nuclear tool porosity minus sonic porosity.

In this problem, nuclear tool porosity is given: 20 pu, and a sonic tool slowness of 79 microseconds/ft over the zone. The known slowness in sandstone & oil (shown in the GB variable box) are 55.5 & 232 microseconds/ft.

Wylie's Equation is next in the GB:
Por sonic = (dt - dtma)/(dtf - dtma) = (79 - 55.5)/(232 - 55.5) = 13 pu.

Since diagenetic = secondary = total - primary: 20 - 13 = 7 pu.

Economics: 2017 #11

Problem 11. Your company invests $100,000 in operating equipment that returns $40,000 annually for 3 years. The interest rate is 5 percent. The equipment has a salvage value of $50,000. The investment’s discounted net cash flow ($ thousand) is closest to:  (A) $142  (B) $152 (C) $42 (D) $52.

Solved in the Guidebook for Net Present Value (NPV).  This problem, however, asks for discounted net cash flow (NCF). So just ignore the initial investment of $100M: answer (B).

This problem is too easy but I include it to help practice reading questions carefully. Note that the NPV answer is included as an option, and since that's a more common calculation it's the kind of mistake can happen to anyone. So always check work.